Question

Find the tenth term in the sequence.

\(a_n = na_{n-1}\)
\(a_1 = -1\)

Answer 1

@jim_thompson5910

Answer 2

can you find a2 first?
there might be a pattern we can find so we don't have to go all the way to a10

Answer 3

to find a2 replace the n's with 2

Answer 4

Maybe..
\(a_2 = 2a_{2-1}\)
\(a_2 = 2(a_1)\)
\(a_2 = 2(-1)\)
\(a_2 = -2\)

Answer 5

that's right

Answer 6

So the common ratio is 2?

Answer 7

no
this isn't a geometric sequence

Answer 8

Why?

Answer 9

geometric sequence would be a_n=constant*a_(n-1)

and here the common ration would be that constant

but n is not a constant
it changes

Answer 10

I learned the geometric sequence formula is like: \(a_n = a_1(r)^{n-1}\). Is it the same thing as what you wrote?

Answer 11

the thing you wrote is an explicit form of a geometric sequence
the thing I wrote is implicit form of a geometric sequence

Answer 12

I didn't learn about the implicit form.

Answer 13

\[a_n=a_1 r^{n-1} \\ \text{ then } a_{n-1}=a_1 r^{(n-1)-1}= a_1 r^{n-2} \\ \text{ and } \frac{a_n}{a_{n-1}}=\frac{a_1 r^{n-1}}{a_1 r^{n-2}} \\ \implies \frac{a_n}{a_{n-1}}=\frac{r^{n-1}}{r^{n-2}} \\ \implies \frac{a_n}{a_{n-1}}=r^{(n-1)-(n-2)} \\ \implies \frac{a_n}{a_{n-1}}= r^{1} \\ \\ \implies a_n= r \cdot a_{n-1}\]

Answer 14

I started with your explicit form for a geometric sequence
and I found a_(n-1) by replacing n with n-1

and then the steps after is on the way to revealing the implicit form
like I divide the a_n thing by the a_(n-1) thing
to show our equations were equivalent

Answer 15

OK thank you for your help :)

Answer 16

no problem :)